Question

Hi . Pls help me with this. Refer pic. ​

Answer 1

Answer:

1+root5, 1-root5

Step-by-step explanation:

Lcm is 2x

So,

2+x/2x=1/x-2

(x+2)*(x-2)=2x

By using identity (a+b)*(a-b) =a^2-b^2

We get,

x^2-4=2x

x^2-2x-4=0

2+-root(4+16)/2. [-b+-root(b^2-4ac)/2a]

2+-2 root 5/2

1+-root 5

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Hope it helps

Answer 2

$\bigstar$ Question:

Solve the given equation for x

• $\frac{1}{x} + \frac{1}{2} = \frac{1}{x \: - \: 2}$

$\bigstar$Given:

The given equation:

• $\frac{1}{x} + \frac{1}{2} = \frac{1}{x - 2}$

$\bigstar$To find:

• The value of x.

$\bigstar$ Solution:

$\frac{1}{x} + \frac{1}{2} = \frac{1}{x - 2} \\ \implies \: \frac{2}{2(x)} + \frac{ x}{2(x)} = \frac{1}{x - 2} \\ \implies \: \frac{2}{2x} + \frac{x}{2x} = \frac{1}{x - 2} \\ \implies \: \frac{2 + x}{2x} = \frac{1}{x - 2}$

Now, here we have to cross - multiply.

$(2 + x)(x - 2) = (2x)(1) \\ \implies \: 2(x) + (2)( - 2) + (x)(x) + (x)( - 2) = 2x\\ \implies \: 2x - 4 + {x}^{2} - 2 = 2x \\ \implies \: 2x - 2x + {x}^{2} - 6 = 0 \\ \implies \: {x}^{2} = 6 \\ \implies \: x = \sqrt{6}$

$\bigstar$ Answer:

• Therefore, the value of x is √6